Question

In young double slit experiment $\frac{d}{D}$= 10^–4 and wavelength of light is used 6000 Å. At a point P on the screen resulting intensity is equal to the intensity due to individual slit I₀. Then the distance of point P from the central maximum is-

• A. 2 mm
• B. 1 mm
• C. 0.5 mm
• D. 4 mm

Answer 1

Answer: (A)

2 mm

Answer 2

I₀ = I₀ + I₀ + 2I₀ cos f

f = 120⁰

f = $\frac{2\pi}{3}$

Dx = $\frac{\lambda}{2\pi}$× $\frac{2\pi}{3}$ = $\frac{\lambda}{3}$

Dx = $\frac{dx}{D}$ = $\frac{\lambda}{3}$

10^–4 x = $\frac{6000 \times 10^{–7}}{3}$

x = 2 mm